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VladimirAG [237]
3 years ago
10

If you had a total sample of 35 candium- 15 M&Mium, 12 Skittlitium and 8 MPeanuttium. What is the relative abundance for eac

h
Chemistry
1 answer:
Flura [38]3 years ago
4 0

<u>Given:</u>

Total Candium = 35

# of M&M = 15

# of Skittlitium = 12

# of Peanutttium = 8

<u>To determine:</u>

The relative abundance  for each

<u>Explanation:</u>

Relative abundance = amount of each item in the sample/Sample total

M&M = 15/35 = 0.429 (or) 42.9 %

Skittlitium = 12/35 = 0.343 (or) 43.3 %

Peanuttium = 8/35 = 0.229 (0r) 22.9%

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Answer:

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3 years ago
If 50.0 g of formic acid (hcho2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of soluti
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If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

<h3>To calculate concentration of sodium formate and formic acid</h3>

The ratio of number of moles and the volume of solution is Molar concentration of substance.

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pH = 3.35

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